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Anna35 [415]
1 year ago
9

Exactly 3.705kg of substance Y are needed to neutralise 100 moles of HCL aq. What could be substance Y ? A. Mgcl2 and Caco3 B. M

gco3 and Ca(NO3)2 C. Mg(NO3)2 and Ca(NO3)2 D. Mgo and Cao
Chemistry
1 answer:
rosijanka [135]1 year ago
5 0

Answer:

C) Ca(OH)2

Note: The options from the question are wrong. The correct options are given below:

15. Exactly 3.705 kg of substance Y are needed to neutralise 100 moles of HCl(aq).

What could be substance Y?

A Ca B CaO C Ca(OH)2 D CaCO3

Explanation:

A) Ca + 2HCl ---> CaCl2 + H2

Molar mass of Ca = 40 g/mol

Mass of Ca required to neutralize 100 moles of HCl = 1/2 * 100 * 40 g = 2000g or 2 kg

B) CaO + 2HCl ---> CaCl2 + H20

Molar mass of CaO = 56.1 g/mol

Mass of CaO required to neutralize 100 moles of HCl = 1/2 * 100 * 56.1 g of CaO = 2805 g or 2.805 kg

C) Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

Molar mass of Ca(OH)2 = 74.1 g/mol

Mass of Ca(OH)2 required to neutralize 100 moles of HCl = 1/2 * 100 * 74.1 g = 3705 g or 3.705 kg

D) CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 required to neutralize 100 moles of HCl = 1/2 * 100 * 100 = 5000 g or 5.00 kg

Therefore, the correct option is C.

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3 0
2 years ago
The laser in a 3D printer uses light with a frequency of 3.844 × 1014 Hz. Calculate the wavelength of this light in nm. Show all
Natali5045456 [20]

Answer:

The wavelength of this light is 780.4 nm.

Explanation:

Given that,

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We need to find the wavelength of this light.

We know that,

c=f\lambda

Where

\lambda is the wavelength of light

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.844 \times 10^{14}}\\\\=7.804\times 10^{-7}\ m\\\\=780.4\ nm

So, the wavelength of this light is 780.4 nm.

5 0
1 year ago
The volume of a gas is 36.0 ml at 10.0°c and 4.50 atm. at what temperature (°c) will the gas have a pressure of 3.50 atm and a v
galben [10]
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
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5 0
1 year ago
A gas is initially at a pressure of 0.43 atm, and a volume of 11.7 liters. Then the pressure is raised to 3.61 atm and the volum
lukranit [14]

Answer:

D. 91.98K

Explanation:

The General Gas Law equation is given by,

\frac{P_1V_1}{T_1}  =  \frac{P_2V_2}{T_2}

From the question,

the initial pressure,

P_1 = 0.43 \: atm

the initial volume,

V_1 = 11.7 \: litres

the final temperature,

T_2=627K

the final pressure,

P_2=3.61atm

the final volume,

V_2=9.5litres

Making

T_1

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T_1= \frac{P_1V_1T_2}{P_2V_2}

By substitution,

T_1= \frac{0.43 \times11.7 \times627}{3.61 \times9.5}

T_1=91.980K

Hence the initial temperature was 91.98 K

4 0
1 year ago
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ArbitrLikvidat [17]
1.22 mg is the same as .022 grams. One gram equals 1000 mg. Convert mg to grams  by dividing mg by 1000mg/gm. Volume = mass over density. Volume equals .022 grams over .754 grams/cm3 which equals .029 or .03cm3. 2.weight in kg times gravity constant. Weight in newtons equals 10kg times 9.8m/s2 that equals 98 newtons. 
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